A276920 Number of solutions to x^3 + y^3 + z^3 + t^3 == 0 (mod n) for 1 <= x, y, z, t <= n.
1, 8, 27, 72, 125, 216, 595, 704, 1539, 1000, 1331, 1944, 3133, 4760, 3375, 5632, 4913, 12312, 8911, 9000, 16065, 10648, 12167, 19008, 16125, 25064, 45927, 42840, 24389, 27000, 35371, 47104, 35937, 39304, 74375, 110808, 58645, 71288, 84591, 88000
Offset: 1
Examples
For n = 3, we see that all nondecreasing solutions {x, y, z, t} are in {{1, 1, 1, 3}, {1, 1, 2, 2}, {1, 2, 3, 3}, {2, 2, 2, 3}, {3, 3, 3, 3}}. The numbers in the sets can be ordered in 4, 6, 12, 4 and 1 ways respectively. Therefore, a(3) = 4 + 6 + 12 + 4 + 1 = 27. - _David A. Corneth_, Oct 11 2016
Links
- Chai Wah Wu, Table of n, a(n) for n = 1..10000 (terms n = 1..242 from Robert Israel)
Programs
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Maple
CF:= table([[false, false, true] = 12, [true, false, false] = 12, [true, false, true] = 6, [false, false, false] = 24, [true, true, true] = 1, [false, true, true] = 4, [false, true, false] = 12, [true, true, false] = 4]): f1:= proc(n) option remember; local count, t, x,y,z,signature; if isprime(n) and n mod 3 = 2 then return n^3 fi; count:= 0; for t from 1 to n do for x from 1 to t do for y from 1 to x do for z from 1 to y do if t^3 + x^3 + y^3 + z^3 mod n = 0 then signature:= map(evalb,[z=y,y=x,x=t]); count:= count + CF[signature]; fi od od od od; count end proc: f:= proc(n) local t; mul(f1(t[1]^t[2]),t=ifactors(n)[2]) end proc: map(f, [$1..40]); # Robert Israel, Oct 13 2016 -
Mathematica
JJJ[4, n, lam] = Sum[If[Mod[a^3 + b^3 + c^3 + d^3, n] == Mod[lam, n], 1, 0], {d, 0, n - 1}, {a, 0, n - 1}, {b, 0, n - 1}, {c, 0 , n - 1}]; Table[JJJ[4, n, 0], {n, 1, 50}] -
PARI
a(n) = sum(x=1, n, sum(y=1, n, sum(z=1, n, sum(t=1, n, Mod(x,n)^3 + Mod(y,n)^3 + Mod(z,n)^3 + Mod(t,n)^3 == 0)))); \\ Michel Marcus, Oct 11 2016
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PARI
qperms(v) = {my(r=1,t); v = vecsort(v); for(i=1,#v-1, if(v[i]==v[i+1], t++, r*=binomial(i, t+1);t=0));r*=binomial(#v,t+1)} a(n) = {my(t=0); forvec(v=vector(4,i,[1,n]), if(sum(i=1,4,Mod(v[i],n)^3)==0, t+=qperms(v)),1);t} \\ David A. Corneth, Oct 11 2016 -
Python
def A276920(n): ndict = {} for i in range(n): i3 = pow(i,3,n) for j in range(i+1): j3 = pow(j,3,n) m = (i3+j3) % n if m in ndict: if i == j: ndict[m] += 1 else: ndict[m] += 2 else: if i == j: ndict[m] = 1 else: ndict[m] = 2 count = 0 for i in ndict: j = (-i) % n if j in ndict: count += ndict[i]*ndict[j] return count # Chai Wah Wu, Jun 06 2017
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